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sveticcg [70]
3 years ago
5

pl help it’s for a grade

Mathematics
1 answer:
Gwar [14]3 years ago
8 0
The answer is b i show the working out in this photo

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Alekssandra [29.7K]

Answer:

2*5*7

Step-by-step explanation:

2*5=10

10*7=70

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Calculate the percent decrease 34,000 2,720
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Take 2,720/34,000=.0806. This means that 2,720 is 8.06% of the original number. In order to find the percentage of decrease, you take 100%-8.06%, which equals 91.94%. Your answer is 91.94%.
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Bart makes $1,000 per month plus his commission on sales. He gets 6 percent of everything he sells. If he sells an $850 mattress
mamaluj [8]
$1,051. You first multiply 850 by 0.06, or 6 percent, to get $51. then add to his monthly salary. Hope this helps
8 0
4 years ago
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twice last month, judy carter rented a car in Fresno, California and traveled around the southwest on business. the car rentel a
mylen [45]

Create a system of 2 equations

Let x be daily fee and y the mileage charge, then

4x + 370y = 244.50...........(1)

3x + 190y = 161.50..............(2)

If we multiply (1) by 3 and (2) by -4 and add the 2 equations we eliminate x:-

12x + 1110y = 733.50

-12x - 760y = -646 Adding :-

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6 0
3 years ago
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Harold uses the binomial theorem to expand the binomial (3x^5 - 1/9y^3)^4
riadik2000 [5.3K]
<h3><em>The complete question:</em></h3>

<u><em> </em></u><u>Harold uses the binomial theorem to expand the binomial </u>(3x^5 -\dfrac{1}{9}y^3)^4<u />

<u>(a)    What is the sum in summation notation that he uses to express the expansion? </u>

<u>(b)    Write the simplified terms of the expansion.</u>

Answer:

(a). (3x^5 -\dfrac{1}{9}y^3)^4=$$\sum_{k=0}^{n}  \binom{4}{k}(3x^5)^{4-k}( -\dfrac{1}{9}y^3)^k $$

(b).(3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561}

Step-by-step explanation:

(a).

The binomial theorem says

(x+y)^n=$$\sum_{k=0}^{n}  \binom{n}{k}x^{n-k}y^k $$

For our binomial this gives

\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=$$\sum_{k=0}^{n}  \binom{4}{k}x^{4-k}y^k $$}

(b).

We simplify the terms of the expansion and get:

$$\sum_{k=0}^{n}  \binom{4}{k}(3x^5)^{4-k}y^k $$= \binom{4}{0}(3x^5)^{4-0}(-\dfrac{1}{9}y^3 )^0+\binom{4}{1}(3x^5)^{4-1}(-\dfrac{1}{9}y^3 )^1+\\\\\binom{4}{2}(3x^5)^{4-2}(-\dfrac{1}{9}y^3 )^2+\binom{4}{3}(3x^5)^{4-3}(-\dfrac{1}{9}y^3 )^3+\binom{4}{4}(3x^5)^{4-4}(-\dfrac{1}{9}y^3 )^4

$$\sum_{k=0}^{n}  \binom{4}{k}(3x^5)^{4-k}(-\frac{1}{9}y^3 )^k $$= (3x^5)^{4}+4(3x^5)^{3}(-\frac{1}{9}y^3 )+6(3x^5)^{2}(-\frac{1}{9}y^3 )^2+\\\\4(3x^5)(-\frac{1}{9}y^3 )^3+(-\frac{1}{9}y^3 )^4

\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561}   }

3 0
3 years ago
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